On a Sobolev type inequality related to the

weighted p-Laplace operator

Marita Gazzini and Roberta Musina

Abstract. In this paper we deal with some Sobolev-type inequalities with weights that were proved by

Mazya in [27]. For integers 1 k < N denote points RN = Rk RNk as pairs (x, y). Let p (1, N),q (p, p] and assume ba := N qNp+ap < k. Then there exists c > 0 such that

c

ZRN

|x|ba |u|q dp/q

ZRN

|x|a|u|p d , u Cc (RN ) .

We prove that the best constant is achieved for any a, p, k, provided that q < p or q = p and a < 0.

Results for weighted Sobolev-type inequalities on cones are also given.

Key Words: Variational methods, critical growth, weighted Lp-Laplace operator, Caffarelli-Kohn-Nirenberg

inequalities, Mazya inequalities.

2000 Mathematics Subject Classification: 35J20, 35J70, 35B33

Ref. SISSA 18/2008/M

Work supported by Cofin. M.I.U.R. Progetto di Ricerca Metodi Variazionali ed Equazioni DifferenzialiNonlineari

SISSA, Via Beirut 4 34010 Trieste, Italy, e-mail: gazzini@sissa.itDipartimento di Matematica ed Informatica, Universita` di Udine, via delle Scienze, 206 33100 Udine,

Italy, e-mail: musina@dimi.uniud.it

1

Introduction

Let k,N be positive integers with 1 k < N . We put RN = Rk RNk, and we denotepoints in RN as pairs (x, y) Rk RNk. Let a, p, q be real parameters, such that

1 < p < N , (pN) kN

< a , max{p ,

p(N k)N p+ a

}< q < p =

pN

N p , (0.1)

and setba = ba(p, q) := N qN p+ a

p. (0.2)

Then there exists a constant c > 0 such that

c

(RN

|x|ba |u|q d)p/q

RN

|x|a|u|p d , u Cc (RN ). (0.3)

Inequality (0.3) was proved by Caffarelli-Kohn-Nirenberg in [11] for spherical weights, andby Mazya, in Section 2.1.6 of [27], in the cylindrical case k < N . In this paper we investigatethe natural question of the existence of extremals for the best constant

Sa,q(p) := infD1,p(RN ;|x|ad)

RN |x|a|u|p d(

RN |x|ba |u|q d)p/q , (0.4)

where the Banach space D1,p(RN ; |x|ad) is the completion of Cc (RN ) with respect tothe norm up = RN |x|a|u|p. Extremals for Sa,q(p) are ground state solutions of theEuler-Lagrange equation

div(|x|a|u|p2u) = |x|ba |u|q2u on RN , (0.5)

which can be regarded as a model for more general degenerate and singular elliptic equations.Several existence result are available in the literature. For a = 0 and q = p the infimum

Sa,q(p) coincides with the Sobolev constant S(p). It is achieved on D1,p(RN ) by an explicitlyknown radially symmetric map (see [31]). In case k = N , Horiuchi has proved existence forq < p or q = p and a < 0 ([23], see also [14] for the case p = 2). Finally, as concerns thecylindrical case k < N we quote [6], where a = 0, k 2 and q (p, p) are assumed, and[32], [28], that deal with p = 2, a 2 k.

Our approach to the minimization problem (0.4) looks quite simple and flexible, and itis uniform with respect to the parameters k,N, a, p and q. It works also for k = N , givingin this way an alternative proof of Horiuchis existence result. Our first main theorem is thefollowing.

Theorem 0.1 Assume that (0.1) is satisfied. Then Sa,p(p) is achieved provided that

q < p or q = p and Sa,p(p) < S(p) .

2

As one can expect, the limiting case q = p is more difficult. It will be shown inProposition A.8 that Sa,p(p) S(p) for any exponent a. The strict inequality holds truewhenever a is negative (see also Proposition A.10 for p = 2, k = 1). This is our second mainresult.

Theorem 0.2 Let p (1, N). Then Sa,p(p) is achieved provided that (pN) kN < a < 0.

Theorem 0.1 extends the existence result in [6] to a = 0, k < p and p(Nk)Np < q < p.

Finally, we notice that Theorem 0.2 gives a positive answer to a question that has beenraised by Tertikas and Tintarev in [32], Section 6, at point 4., at least when p < k.

Let us describe the main features of problem (0.3) and our approach. While study-ing (0.4) one has to take into account the action of the groups of dilations in RN and oftranslations in RNk. Indeed, for any minimizing sequence uh, and for arbitrary sequencesth (0,+), yh RNk, it turns out that uh(x, y) := uh(thx, thy + yh) still approachesthe infimum Sa,q(p). By this remark it is quite easy to exhibit non compact minimizingsequences. The group of translations in the x-variable produces worse lack of compactnessphenomena in the limiting case q = p, since minimizing sequences for (0.4) might blow-upan extremal for the Sobolev constant S(p).

Our main idea to prove Theorem 0.1 simply consists in looking for a good minimizingsequence. This strategy has been already followed in [28]. The main tools are the Ekelandvariational principle, a Rellich-type theorem, and a suitable rescaling argument. In par-ticular, we skip the blow-up analysis of all minimizing sequences (as it has been done forexample in [14], [6] and in [32]). We dont require the Brezis-Lieb Lemma, the Concentration- Compactness Lemmata by P.L. Lions, as well as any similar result.

The crucial step is Proposition 1.3. It deals with the asymptotic behavior of boundedsequences of approximate solutions to (0.5). Thanks to Proposition 1.3, we can find a weaklyconvergent minimizing sequence uh whose Lq-norms are bounded away from 0 on a compactsubset of (Rk\{0})RNk. In this way we exclude concentration at 0 and vanishing, and weovercome the lack of compactness produced by dilations in RN and by translations in RNk.If q < p we use the Rellich Theorem and we conclude that uh converges weakly to someu 6= 0. Then standard arguments imply that u achieves Sa,q(p). If q = p the assumptionSa,p(p) < S(p) prevents concentration phenomena at points (x0, y0) with x0 6= 0, and wecan conclude as in the subcritical case.

There is a large number of papers that are related to (0.3). Besides the above quotedpapers we cite for example [1], [2], [3], [4], [5], [7], [9] [12], [16], [17], [18], [21], [22], [29], [30],[34] for k = N , and [8], [13], [19], [24], [25], [26], [33] for k < N .

The paper is organized as follows:

3

in Sections 1 and 2 we prove our main Theorems; in Section 3 we establish a suitable Sobolev type inequality on cones for any a R,

a 6= p k (compare with Lemma 3.1), and we discuss the existence of extremal functions.The main result is stated in Theorem 3.2;

in an Appendix we collect additional results, remarks and open problems. In particularwe draw here our attention on the differential equation (0.5) and on the limiting case q = p.

Notation For any integer j 1, we denote by BjR(z) the j-dimensional ball of radius R centeredat z Rj . The Lebesgue measure of a domain in RN is denoted by ||.

Let q [1,+), R, and let be a domain in RN . Then Lq(; |x|d) is the space ofmeasurable maps u on with

R|x||u|q d < +, so that Lq(; |x|0d) Lq() is the standard

Lebesgue space. For p > 1 the reflexive Banach space D1,p() is the closure of Cc () with respectto the Lp-norm of u. The Sobolev critical exponent is p = Np

Np . We recall that the best constant

S(p) := S0,p(p) = infD1,p(RN )

RRN |u|p d`R

RN |u|p dp/p

is achieved on D1,p(RN ) by the map

U() :=1 + || pp1

Npp

.

1 Proof of Theorem 0.1

We start with two technical lemmata. The first one is a Rellich-type result.

Lemma 1.1 Let RN be a bounded domain. Then D1,p(RN ; |x|ad) Lp(, |x|ad)with compact inclusion.

Proof. Fix a map u Cc (RN ). Holder inequality and (0.3) give

|x|a|u|p d || pN(

|x| NaNp |u|p d)p/p

c|| pNRN

|x|a|u|p d , (1.1)

where c does not depend on u. This proves the continuity of the embedding. To prove com-pactness take a sequence uh in D1,p(RN ; |x|ad), with uh 0 weakly in D1,p(RN ; |x|ad).Fix > 0 and take a smooth function C(Rk) such that 0 1, (x) = 0 if|x| 2, and (x) = 1 if |x| . By Rellich Theorem, turns out that

|x|a|uh|p d = o(1)

as h +, since |x| is bounded away from 0 on the support of . On the other hand, thesequence uh is bounded in D1,p(RN ; |x|ad), and therefore from (1.1) one gets

|x|a|(1 )uh|p d c||pN

RN

|x|a|uh|p d c||pN ,

4

where := {(x, y) | |x| < }. Writing uh = uh + (1 )uh one infers that

|x|a|uh|p d c

|x|a(|uh|p + |(1 )uh|p) d o(1) + c||pN

for fixed, as h +. The conclusion easily follows, since || 0 as 0.

Lemma 1.2 If Cc (RN ) then u D1,p(RN ; |x|ad) for any u D1,p(RN ; |x|ad).

Proof. We can approximate any fixed u D1,p(RN ; |x|ad) with a sequence uh Cc (RN ).Using Lemma 1.1 it is easy to prove that uh u in D1,p(RN ; |x|ad), hence u D1,p(RN ; |x|ad).The main step in the proof of Theorem 0.1 is the following Proposition, that is concernedwith approximate solutions to (0.5).

Proposition 1.3 Let uh be a bounded sequence in D1,p(RN ; |x|ad), and let fh 0 be asequence in the dual of D1,p(RN ; |x|ad). Assume that for a, p, q, ba as in (0.1), (0.2) itholds that

div(|x|a|uh|p2uh) = |x|ba |uh|q2uh + fh .Then, up to a subsequence, either uh 0 strongly in Lq(RN ; |x|bad), or there existsequences (th)h (0,+) and (h)h RNk, such that

limh+

K

|x|ba |uh|q d > 0 ,

where uh(x, y) = tNp+a

p

h uh(thx, thy + h) and K = {(x, y) RN | 12 < |x| < 1 , |y| < 1 }.

Proof. We can assume that there exists u D1,p(RN ; |x|ad) such that uh u weaklyin D1,p(RN ; |x|ad) and in Lq(RN ; |x|bad). Notice that u solves (0.5). If u 6= 0 thenwe are done, since by the maximum principle u cannot vanish on K. Then the conclusionfollows by the weak lower semicontinuity of the Lq-norm. Therefore, we assume u = 0 andlimh+

RN |x|ba |uh|q d > 0. Fix 0 > 0 in such a way that

q

qp0 < lim

h+

RN

|x|ba |uh|q d , 20 < Sa,q(p) .

Using in a standard way the concentration function

Qh(t) := supRNk

Bkt (0)BNkt ()

|x|ba |uh|q d ,

it is possible to select th > 0 and h RNk such that the rescaled sequence

uh(x, y) := tNp+a

p

h uh(thx, thy + h)

5

satisfies |x|a|uh|p = |x|a|uh|p = O(1), and

Bk1 (0)BNk1 (y)|x|ba |uh|q d (20)

qqp y RNk , (1.2)

Bk1 (0)BNk1 (0)|x|ba |uh|q d

qqp0 > 0 , (1.3)

div(|x|a|uh|p2uh) = |x|ba |uh|q2uh + fh , (1.4)

with fh 0 in D1,p(RN ; |x|ad). As before, if (up to a subsequence) uh u 6= 0 then weare done. If uh 0, choose a finite number of points y1, ..., ys RNk such that

BNk1 (0) s

j=1

BNk1/2 (yj) . (1.5)

Let 1, ...s be cut-off functions, with j = j(y) Cc (BNk1 (yj)), j 1 on BNk1/2 (yj)and 0 j 1. Also, fix a map = (x) Cc (Bk1 (0)) satisfying 0 1 and 1 onBk1/2(0). Thanks to Lemma 1.2 we can use

ppi uh as test function in (1.4) to findRN

|x|a|uh|p2uh (ppj uh) d =RN

|x|ba |uh|qp|j uh|p d + o(1) . (1.6)

Direct computations and Lemma 1.1 giveRN

|x|a|uh|p2uh (ppj uh) d =RN

|x|a|(uh)|p d + o(1).

Thus, we can use Holder inequality, (1.2), (1.6) and the definition of Sa,q(p) to infer that

Sa,q(p)(

RN|x|ba |j uh|q d

) pq

20(

RN|x|ba |j uh|q d

) pq

+ o(1) .

Since 20 < Sa,q(p) this implies thatRN |x|ba |j uh|q d = o(1) , and therefore, by (1.5),

Bk1/2(0)BNk1 (0)|x|ba |uh|q d

sj=1

Bk1/2(0)BNk1/2 (yj)

|x|ba |uh|q d = o(1) .

Finally, from (1.3) we get

0 < q

qp0 0 , (1.9)

where K = {(x, y) Rk RNk | 12 < |x| < 1 , |y| < 1 }. We claim that u 6= 0. Thisis immediate if q < p, since in this case

K|x|ba |u|q d = limh+

K|x|ba |uh|q d > 0

by Rellich Theorem. Therefore we take q = p. Choose smooth maps Cc (Rk) and Cc (RNk) in such a way that (x) = 0 for |x| 14 , (x) = 1 for 12 |x| 1 and(y) = 1 for |y| 1. Notice that 1 on K. Since fh, ppuh = o(1), we can argue asin the proof of Proposition 1.3 to get

RN|x|a|(uh)|p d Sa,p(p)

(RN

|x| NaNp |uh|p d) p

p

+ o(1) . (1.10)

Now, notice that |x| ap(uh) = (|x| apuh) Fh, where Fh := uh(|x| ap ). Since has compact support and since it vanishes in a neighborhood of the singular set {x = 0},then Fh 0 in Lp(RN )N by Rellich Theorem. Therefore

RN|x|a|(uh)|p d =

RN

|(|x| apuh)|p d + o(1)

S(p)(

RN|x| NaNp |uh|p d

) pp

+ o(1)

by Sobolev inequality. In this way from (1.10) we get

S(p)(

RN|x| NaNp |uh|p d

) pp

Sa,p(p)(

RN|x| NaNp |uh|p d

) pp

+ o(1).

Since Sa,p(p) < S(p) by assumption, this implies thatK

|x| NaNp |uh|p d RN

|x| NaNp |uh|p d = o(1) ,

that contradicts (1.9). Thus, uh u 6= 0 weakly in D1,p(RN ; |x|ad).

7

Finally, standard arguments imply that uh u strongly in D1,p(RN ; |x|ad), and there-fore that u achieves Sa,q(p). For completeness we recall the argument here. From (1.8) itfollows that u solves (0.5), and in particular

RN|x|a|u|p =

RN

|x|ba |u|q (Sa,q(p))qp

(RN

|x|a|u|p) q

p

,

by definition of Sa,q(p). Since u 6= 0, this implies thatRN |x|a|u|p (Sa,q(p))

qqp . But

then, (1.7) and the lower semicontinuity of the norm in D1,p(RN ; |x|ad) implyRN

|x|a|uh|p =RN

|x|a|u|p + o(1) ,

that suffices to conclude that uh u in the uniformly convex space D1,p(RN ; |x|ad).

2 Proof of Theorem 0.2

In order to prove Theorem 0.2 it suffices to show that Sa,p(p) < S(p). Indeed, we claimthat the following estimate holds:

Sa,p(p) S(p) N pN

N + aN p a(p 1) . (2.1)

Notice that the right hand side in (2.1) is strictly increasing in a, and therefore (2.1) impliesSa,p(p) < S(p) for a < 0. To prove (2.1) we estimate Sa,q(p) with the map

U() =(1 + || pp1

)Npp,

that achieves the best constant S(p) (see [31]). We compute

|U |p =(N pp 1

)p|| pp1N , |U |p = N , (2.2)

where we have set () := 1 + || pp1 . An application of the divergence theorem leads toRN|x|a|| pp1N = p 1

p

1N 1

RN|x|a(1N ) = p 1

p

N + aN 1

RN|x|a1N .

On the other hand,RN|x|a|| pp1N d =

RN|x|a (1N N) d ,

and hence RN|x|a|| pp1N d = (p 1)(N + a)

N p a(p 1)RN|x|aN d .

8

Thus, from (2.2) we inferRN|x|a|U |p d =

(N pp 1

)p (p 1)(N + a)N p a(p 1)

RN|x|aN d . (2.3)

We can compute S(p) by setting a = 0 in (2.3):

S(p) =(N pp 1

)p (p 1)NN p

(RN

N d) p

N

,

Therefore, (2.3) and Holder inequality implyRN|x|a|U |p

(N pp 1

)p (p 1)(N + a)N p a(p 1)

(RN|x| NaNpN

)NpN

(RN

N) p

N

= S(p)N pN

N + aN p a(p 1)

(RN|x| NaNpUp d

)NpN

,

and the conclusion readily follows.

Remark 2.1 We notice that the condition p2 < N suggested in [32] to get the existence ofa minimizer for Spk,p(p) is not necessary, even if up to now we are not able to prove itssufficiency (except when p = 2, compare with [32]).

3 Problems on cones

In this section we extend some results already proved in [12] and in the more recent papers[7], [28].

Our arguments for Theorem 0.1 can be used with no modifications to study problemson cones. Accordingly with [12], we say that a cone in Rk is a domain Ck Rk such thatx Ck for every > 0 and for every x Ck. A cone Ck is said to be proper if 0 / Ck.Notice that Rk itself is a cone, Rk \ {0} is a proper cone, and that (0,+) is a proper conein R. The only domains in RN that are invariant with respect to dilations and translationsin the y-variable are of the form Ck RNk, where Ck is a cone.

In this section we take = Ck RNk, with Ck a proper cone in RN . Notice that forany a R the Hardy inequality holds on :k p+ ap

p

|x|ap|u|p d

|x|a|u|p d u Cc () ,

see for example [15]. If a 6= pk the Hardy constant is positive, and we can define the Banachspace D1,p0 (; |x|ad) by completing Cc () with respect to the norm

|x|a|u|p. Notice

that D1,p0 (; |x|ad) D1,p(RN ; |x|ad) for a > (pN) kN . By a density argument based on

9

the Hardy inequality one can prove that D1,p0 (Rk \ {0})RNk; |x|ad) = D1,p(RN ; |x|ad)if and only if a > p k.

We start with a Sobolev type inequality on cones.

Lemma 3.1 Assume p (1, N), q [p, p], a 6= p k and set ba = N qNp+ap . Let = Ck RNk, with Ck a proper cone in Rk. Then there exists a constant c > 0 such that

c

(

|x|ba |u|q d)p/q

|x|a|u|p d , u D1,p0 (; |x|ad). (3.1)

Proof. Notice that (3.1) is the Hardy inequality if q = p. We are going to prove (3.1) forq = p. Fix any map u Cc (), and set Lu := |x|

ap u Cc (). Notice that

|Lu|pd c(

|x|a|u|p d +

|x|ap|u|p d) c

|x|a|u|p d

where the constants c dont depend on u. Thus,(

|x| NaNp |u|p d) p

p

=(

|Lu|p d) p

p

c

|x|a|u|p d

by Sobolev inequality. Then (3.1) for q = p follows by a density argument. For q (p, p)inequality (3.1) can be proved by interpolating between the cases q = p and q = p, viaHolder inequality.

Thanks to Lemma 3.1, for any a 6= p k, q (p, p] the infimum

Sa,q(p; ) := infD1,p0 (;|x|ad)

|x|a|u|p d(

|x|ba |u|q d)p/q

is positive. Notice that Sa,q(p; ) Sa,q(p) if (0.1) is satisfied. One can argue as forTheorem 0.1 to prove the next result. We omit the details.

Theorem 3.2 Let p (1, N), q (p, p], a 6= p k and let = Ck RNk, with Ck aproper cone in Rk. Then Sa,q(p; ) is achieved provided that

q < p or q = p and Sa,p(p; ) < S(p).

Remark 3.3 As usual, the case q = p is more difficult, and up to now we are not ableto give general sufficient conditions for the strict inequality Sa,p(p; ) < S(p). Somethingcan be said in case k = 1, p = 2 and = RN+ := (0,+) RN1. Indeed, if in additionN 4 it has been proved in [28] that Sa,2(2;RN+ ) < S if and only if 0 < a < 1. On thecontrary, it turns out that Sa,6(2;R3+) = S and it is never achieved, as a consequence of thenonexistence result in [26] (the proof goes as for 3. of Proposition A.10 below).

10

Remark 3.4 Let u be an extremal for Sa,q(p; ). Then u is a nonnegative solution to (0.5)on . For example, set x = (x1, ..., xk) Rk and take = (0,+) Rk1 RNk. Thus, is an half-space whose boundary contains the singular set {x = 0}. Theorem 3.2 givessufficient conditions for the existence of positive entire solutions to the Dirichlet problem{div(|x|a|u|p2u) = |x|bauq1 on x1 > 0

u = 0 on {x1 = 0} .

A Appendix

In this Appendix we collect additional results and remarks about the Euler-Lagrange equa-tions related to the Mazya inequality. Some of them are nowadays standard or alreadypartially known, and can be found in literature (for example, in [6] for a = 0 and in [32], [28]for p = 2, a 2k). However, we are going to outline their proofs for sake of completeness.

First of all we notice that Theorems 0.1 , 0.2 provide sufficient conditions for the existenceof non trivial weak entire solutions to{div(|x|a|u|p2u) = |x|ba uq1 in RN ,

u 0 . (A.1)

The argument is well known, and we omit it.

Remark A.1 Notice that if k 2 then u > 0 on {x 6= 0}, by the maximum principle. Thisis no longer true in general if k = 1 (compare with Section A.1).

Remark A.2 Assume k 2 and p = 2. Then every extremal for Sa,q(2) is a weak entiresolution to {div(|x|au) = |x|ba uq1 in RN

u > 0 .(A.2)

It has been proved in [19], Corollary 5.1, that for a 2 k and

max{2,

2(N k)N 2 + a

}< q 2

. Actually, we can get existence since the true critical exponenton the class of symmetric maps is 2(Nk+1)Nk1 .

In addition, if a [2(2 k), 0], every classical positive solution u Lq(RN ; |x|bad) to

div(|x|au) = |x|ba uq1 on {x 6= 0}

is radially symmetric in the x-variable ([19], Corollary 5.2). Finally, breaking symmetryoccurs as a >> 0, that is, minima of (0.4) are not symmetric in the x-variable (see [19],Corollary 5.3).

11

It would be of interest to know whether entire solutions to (A.1) are radially symmetricin the x-variable or not, at least for p < k and a close to p k. Also, one may wonder ifbreaking symmetry occurs as a +, when q (p, p).

We point out a simple technical Lemma that will be useful later-on.

Lemma A.3 Let a p k. Then Cc ((Rk \ {0}) RNk) is dense in D1,p(RN ; |x|ad).

Proof. Fix any map v Cc (RN ). For > 0 set

(|x|) =0 if |x| 2log |x|/2| log | if

2 < |x| < 1 if |x| .

It is clear that v D1,p(RN ; |x|ad) and that (v v) = (1 )v v 0 a.e.on RN , as 0. To prove that v v in D1,p(RN ; |x|ad) it suffices to remark that

RN|x|a|v|p d cv

Rk|x|a||p dx cv| log |1p

since a p k, where the constants cv dont depend on . The conclusion follows viaLebesgues theorem, since |(1 )v| |v| on RN , and since |x|a|v|p L1(RN ).

Remark A.4 Set RN0 := (Rk \ {0}) RNk. Since Rk \ {0} is a proper cone in Rk, thenthe results in Section 3 apply with = RN0 . In particular, Sa,q(p;RN0 ) is achieved forany a 6= p k, q (p, p); in the limiting case Sa,p(p;RN0 ) is achieved provided thatSa,p(p;RN0 ) < S(p). It can be easily proved via Hardy inequality that for a > (pN) kN ,

D1,p0 (RN0 ; |x|ad) = D1,p(RN ; |x|ad) Lp(RN ; |x|apd) .

Hence, D1,p0 (RN0 ; |x|ad) is a proper subspace of D1,p(RN ; |x|ad) if and only if a < p k.In [20] the writing authors compare the infimum Sa,q(p;RN0 ) with Sa,q(p) for a < p k.

In particular, if p = 2 and Sa,q(2;RN0 ) > Sa,q(2) > 0, any extremal for Sa,q(2) corresponds,up to a functional change, to a positive classical solution v Lq(RN ; |x|b0d) to

v |x|2 v = |x|b0 vq1 on {x 6= 0} ,

that is singular on the N k > 0 dimensional subspace {x = 0}.

Remark A.5 In this remark we take k 2. Let Ck be a cone, properly contained inRk \ {0}. Assume that (0.1) are satisfied and that q < p or Sa,p(p; ) < S(p). Then boththe infima Sa,q(p; ) and Sa,p(p) are achieved. One can write down the Euler-Lagrangeequations and use the maximum principle to infer that Sa,q(p) < Sa,q(p; ). This is nolonger true if k = 1 and a p 1, compare with Section A.1 below.

12

A.1 The case k = 1

When k = 1 the singular set {x = 0} is an hyperplane that disconnects the domain into twoproper cones. Let us point out an immediate corollary to Theorem 0.1.

Corollary A.6 Let k = 1, p (1, N) and p(N1)Np < q < p. Then problem{pu = |x|ba uq1 in RNu 0 ,

has a weak entire ground state solution.

As observed in [19], in case p = 2 the solution of Corollary A.6 is even in the x-variable anddecreasing for x > 0. In particular, u can never vanish on RN . This remark and the nextLemma underline the contrast between the cases a = 0 < p 1 and a p 1.

Lemma A.7 Let k = 1, p (1, N), q (p, p] and a p 1. Then every minimizer forSa,q(p) vanishes on a half-plane.

Proof. Set RN := (, 0)RN1 and RN+ := (0,+)RN1. By Lemma A.3, there existsequences uh Cc (RN ) and u+h Cc (RN+ ) such that uh + u+h u in D1,p(RN ; |x|ad).Then

RN|x|a|uh |p

RN

|x|a|u|p ,RN+

|x|a|u+h |p RN+

|x|a|u+h |p ,

and similarly for the weighted Lq norms. Since u and u+ have disjoint supports, then

Sa,q(p) =

RN |x|a|(uh + u+h )|p(RN |x|b|uh + u+h |q

)p/q + o(1) Sa,q(p)

(RN

|x|b|uh |q)p/q

+(

RN+|x|b|u+h |q

)p/q(

RN|x|b|uh |q +

RN+

|x|b|u+h |q)p/q + o(1)

by the Mazya inequality. The conclusion easily follows by letting h +, since p < q.By Lemma A.7 it turns out that Sa,q(p;RN+ ) = Sa,q(p) whenever a p 1, even if both

the infima are achieved. This means that the maximum principle fails in this case. Wesuspect that this is not longer true for a below p 1, as the case p = 2, a = 0 suggests.

A.2 The limiting case q = p

We first point out a first consequence of the action of translations in the x-variable in thelimiting case q = p.

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Proposition A.8 Let 1 k N , 1 < p < N and a > (pN) kN . Then Sa,p(p) S(p).

Proof. Fix u Cc (BN1 (0)) and > 0. Fix a point x0 Rk with |x0| = 1 and set0 = (x0, 0). Set u() := u(1( 0)) Cc (BN (0)). We estimate

Sa,p(p) RN |x|a|u|p d(

RN |x|NaNp |u|p d

)p/q (1 + 1 )|a|

RN |u|p d(RN |u|p d

)p/q ,that is,

Sa,p(p) infCc (B

N1 (0))

RN |u|p d(RN |u|p d

)p/q = S(p) ,by the invariance of the ratio

(RN |u|p

) (RN |u|p

)p/qwith respect to dilations.

Remark A.9 Quite reasonably it happens that Sa,p(p) = S(p) for a large enough. On theother hand, one might suspect that Sa,p(p) < S(p) for a close to p k and k < p

N .

This is the case when p = 2 (see [14] for k = N and compare with Proposition A.10 belowfor k < N).

In case p = 2 we can improve improve Theorem 0.2. In order to simplify notation we setSa := Sa,2(2) and S := S(2), so that Sa S by Proposition A.8.

Proposition A.10 Let 1 k < N , 1 < p < N and a > (2N) kN .

1. Let k 2. Then Sa is achieved if and only if a 0 .

2. Let k = 1 and N 4. Then Sa is achieved if and only if a < 2.

3. Let k = 1 and N = 3. Then Sa is achieved if a 0 and it is not achieved if a 1.

Proof of 1. Immediate, from Theorem 0.2 and from Theorem B.5 of [28].

Proof of 2. Necessity follows again from [28], Theorem B.5. Sufficiency for a < 0 followsfrom Theorem 0.2. Thus, we only have to show that Sa is achieved if a (0, 2), This fact wasalready noticed in [32] for a = 1; we outline here a simpler proof for completeness. To provethat Sa < S for a (0, 2) fix r,R > 0 and take any bounded domain (r,R) RN1.Fix any map v Cc (). Then the integration by parts implies that

RN|x|a|(xa/2v)|2 =

|v|2 a(2 a)4

x2v2

|v|2 a(2 a)4R2

v2 .

Thus

Sa,2(2) infCc ()

|v|2 a(2a)4R2

v2(

|v|2) 22 < S

14

since N 4, by a well known result by Brezis and Nirenberg [10], Lemma 1.1.Proof of 3. By Theorem 0.2, Sa is achieved provided that a < 0. By contradiction, assumethat for some a 1 the infimum Sa is achieved by a map u D1,2(R3; |x|ad). By LemmaA.7 we can assume that u is a positive entire solution to{div(|x|au) = |x|ba u5 in (0,+) R2 ,

u = 0 on {0} R2 .

Set v := xa/2u. Then Lemma A.3 and direct computations imply that v is an entire positivesolution (in the sense of [26]) to{

v = a(2a)4 |x|2v + v5 on (0,+) R2v = 0 on {0} R2 .

This contradicts the nonexistence result in [26], Section 6 (see also [8] for a = 1).

Up to now we dont know whether Sa is achieved or not if k = 1, N = 3 and a (0, 1).This is the only case left open.

Theorems 0.1, 0.2 and Proposition A.10 give an alternative proof of a result by Tertikasand Tintarev [32] in case a = 2 k and p = 2. However, as noticed in Section 6 of [32], if inaddition k > 2 one can use the symmetry of minima to (0.4) (compare with Remark A.2)and a functional change to reduce the proof of existence in this case to an existence resultby Badiale and Tarantello [6].

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